6 g of a substance is dissolved in 100 g of water depresses the freezing point by `0.93^(@)C`. The molecular mass of the substance will be: (`K_(f)` for water ` = 1.86^(@)C`/molal)

Options

A. 60 g/mole
B. 180 g/mole
C. 120 g/mole
D. 240 g/mole

Answer

To find the molecular mass of the substance, we can use the formula for freezing point depression:

\[ \Delta T_f = K_f \cdot m \]

Where:
- \(\Delta T_f\) = depression in freezing point (in °C)
- \(K_f\) = freezing point depression constant (for water, \(K_f = 1.86 \, °C/\text{molal}\))
- \(m\) = molality of the solution

### Step 1: Calculate the molality (m) of the solution

Molality (m) is defined as the number of moles of solute per kilogram of solvent.

Given:
- Mass of the solute = 6 g
- Mass of the solvent (water) = 100 g = 0.1 kg

First, we need to calculate the moles of the solute using the formula:

\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]

### Step 2: Rearranging the freezing point depression formula

From the freezing point depression formula, we can express molality as:

\[ m = \frac{\Delta T_f}{K_f} \]

Substituting the values:

\[ m = \frac{0.93}{1.86} = 0.5 \, \text{molal} \]

### Step 3: Calculate the number of moles of solute

Now, using the definition of molality:

\[ 0.5 = \frac{\text{moles of solute}}{0.1} \]

Rearranging gives:

\[ \text{moles of solute} = 0.5 \times 0.1 = 0.05 \, \text{moles} \]

### Step 4: Calculate the molecular mass of the solute

The molecular mass (M) can be calculated using the formula:

\[ M = \frac{\text{mass of solute}}{\text{moles of solute}} \]

Substituting the values:

\[ M = \frac{6 \, \text{g}}{0.05 \, \text{moles}} = 120 \, \text{g/mol} \]

### Final Answer

The molecular mass of the substance is **120 g/mol**.

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